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Ex(t): = ect (¹+¹). 28(t+2) - 28(t-2) - rect( rect (¹¹) +rect

Ex(t): = ect (¹+¹). 28(t+2) - 28(t-2) - rect( rect (¹¹) +rect

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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|x (t) - - - = 26(t + 2) — 28(t − 2) — rect(+¹) +rect (¹¹) So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)... ..... equation 1. F₁ (w) Fit replace by F₁w F₁ (1) (1) replace by F₂ (w) F2 Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁(t) = −6(t + 2) + 6(t - 2)........ equation 3 now t replace by w, F₁(w) = -2 [+] x 2 F₁(w) F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) = |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw How did he get the value:
Transcribed Image Text:|x (t) - - - = 26(t + 2) — 28(t − 2) — rect(+¹) +rect (¹¹) So, dF(t) dt F2 · = F₁ (t) + F₂ (t) 2 (+)... ..... equation 1. F₁ (w) Fit replace by F₁w F₁ (1) (1) replace by F₂ (w) F2 Apply Fourier transform, jw F(w) = F₁ (w) + F2₂(w)...... equation 2 F₁(t) = −6(t + 2) + 6(t - 2)........ equation 3 now t replace by w, F₁(w) = -2 [+] x 2 F₁(w) F1(w) = 4jwSin²w. and F2(w) = 2 x 4Sin (2) = |F2(w) = 8Sin (2) F2(w) = -8Sin2w jw How did he get the value:
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